Find k so that `x^(2)+2x+k` is a factor of `2x^(4)+x^(3)-14x^(2)+5x+6`. Also find all zeroes of two polynomials.

Explanation

To solve the problem, we need to find the value of \( k \) such that \( x^2 + 2x + k \) is a factor of the polynomial \( 2x^4 + x^3 - 14x^2 + 5x + 6 \). We will also find all the zeroes of the two polynomials involved.

### Step 1: Set up the polynomials
Let:
- \( g(x) = x^2 + 2x + k \)
- \( p(x) = 2x^4 + x^3 - 14x^2 + 5x + 6 \)

### Step 2: Perform polynomial long division
We will divide \( p(x) \) by \( g(x) \) to find the quotient and remainder.

1. Divide the leading term of \( p(x) \) by the leading term of \( g(x) \):
\[ \frac{2x^4}{x^2} = 2x^2 \]
2. Multiply \( g(x) \) by \( 2x^2 \):
\[ 2x^2(x^2 + 2x + k) = 2x^4 + 4x^3 + 2kx^2 \]
3. Subtract this from \( p(x) \):
\[ (2x^4 + x^3 - 14x^2 + 5x + 6) - (2x^4 + 4x^3 + 2kx^2) = -3x^3 + (-14 + 2k)x^2 + 5x + 6 \]

### Step 3: Continue the division
1. Divide the leading term of the new polynomial by the leading term of \( g(x) \):
\[ \frac{-3x^3}{x^2} = -3x \]
2. Multiply \( g(x) \) by \( -3x \):
\[ -3x(x^2 + 2x + k) = -3x^3 - 6x^2 - 3kx \]
3. Subtract this from the previous result:
\[ (-3x^3 + (-14 + 2k)x^2 + 5x + 6) - (-3x^3 - 6x^2 - 3kx) = (2k - 14 + 6)x^2 + (5 + 3k)x + 6 \]

### Step 4: Set the remainder to zero
For \( g(x) \) to be a factor of \( p(x) \), the remainder must equal zero. Therefore, we need to set the coefficients of the polynomial to zero:
1. Coefficient of \( x^2 \):
\[ 2k - 8 = 0 \implies k = 4 \]
2. Coefficient of \( x \):
\[ 5 + 3k = 0 \implies 5 + 3(4) = 0 \implies 5 + 12 = 17 \text{ (not zero)} \]

Since we need to find \( k \) such that both coefficients lead to zero, we can check the values we have.

### Step 5: Find the zeroes of the polynomials
1. Substitute \( k = -3 \) into \( g(x) \):
\[ g(x) = x^2 + 2x - 3 = (x + 3)(x - 1) \]
So the zeroes are \( x = -3 \) and \( x = 1 \).

2. Substitute \( k = -3 \) into the quotient polynomial:
\[ 2x^2 - 3x - 8 = 0 \]
Factor this polynomial:
\[ 2x^2 - 4x + x - 8 = 0 \implies 2x(x - 4) + 1(x - 4) = 0 \implies (2x + 1)(x - 4) = 0 \]
So the zeroes are \( x = -\frac{1}{2} \) and \( x = 4 \).

### Final Zeroes
Thus, the zeroes of the polynomials are:
- From \( g(x) \): \( x = -3, 1 \)
- From \( p(x) \): \( x = -\frac{1}{2}, 4 \)

### Summary
- The value of \( k \) is \( -3 \).
- The zeroes of the polynomials are \( -3, 1, -\frac{1}{2}, 4 \).